How do you solve for the equation of the tangent plane of xy = 2lnz at (2, 0, 1)

xy = 2 ln(z)
xy − 2 ln(z) = 0

Let F(x,y,z) = xy − 2 ln(z)

We find ∇F at point (2, 0, 1)
This will give us vector that is normal to plane at point (2, 0, 1)

∇F = < ∂F/∂x, ∂F/∂y, ∂F/∂z >
∇F = < y, x, −2/z >

∇F(2, 0, 1) = < 0, 2, −2 >

Tangent plane has normal < 0, 2, −2 > and passes through point (2, 0, 1)

0 (x − 2) + 2 (y − 0) − 2 (z − 1) = 0
2y − 2z + 2 = 0
2y − 2z = −2